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Is it feasible to circumnavigate the Earth in a sailplane?
Would it be feasible and survivable to circumnavigate the Earth in a glider or sailplane without propulsion and without landing until the circumnavigation is completed, when you'd return to the same location you were disattached from the carrier plane or winch? What would the major obstacles be and how could one handle it? Did anyone ever attempt such endeavor?
Edit: The two-seated Perlan II sailplane once reached an altitude of more than 70,000 ft (21 km) so perhaps let's discuss more over that single plane, whether the Perlan II, assuming it reaches 70,000 ft, could cross huge areas of sea for instance. If the two pilots manage their time so that one of them navigates while the other sleeps, isn't it likely you could circumnavigate the Earth in the Perlan  II?

The wandering albatross does this using "dynamic soaring", catching air currents that form over the top of ocean waves.  They can also flap, but in practice they don't. They can travel 600 miles a day.  Not an answer, because an albatross isn't legally a sailplane (FAA 21.17-2a). https://jeb.biologists.org/content/221/1/jeb169938

@JamesK Not just legally, it's no plane whatsoever, it's a living entity. Interesting however. Maybe if one could ride a griffin or great eagle one would wonder how long that griffin can soar without flapping and if one could circumnavigate your planet on it.

Humm... Does your circumnavigation have to be east-west?  If you go approximately polewise, you have a lot less ocean to cross.  North & South America should be fairly easy, just follow the Rockies & Andes.  Then if you can get across Antarctica and the stretch of ocean to the southern tip of Africa, you could follow its east coast, across Asia Minor and then north of the Himalayas and cross Bering Strait...

I like where @jamesqf  is going.  Following the updraft from a continental divide should give a way to sail for a very long time.  To be considered a circumnavigation, at least by oceangoing standards, one must cross the equator twice.  One would also have to pass over a pair of antipodes, and travel a distance equal or greater than the equatorial distance.  Following mountain ridges for the updraft makes two of those "easy" but not the requirement of antipodes as there's a lot of water without a ridge to ride.

I like jamesqf's suggestion too. I think it qualifies for a circumnavigation because it crosses and re-crosses the equator and you return to where you began. If james posts this as an answer I'd accept that one.

Clealy possible as with a high enough lunch from a carrier plane, the glider is a satellite.

@IanRingrose Satellites orbit without propulsion but also without aerodynamic lift. This lets me wonder whether it would qualify as an orbit if a sailplane fully circumnavigates the Earth relative to the stars, without propulsion. Actually many satellites experience atmospheric drag so the question is where one would put the difference between an orbit and...something different.

&Giovanni: Not an answer because I don't know whether it's actually feasible.  I don't know enough about Antarctic weather to guess whether there'd be sufficient lift over the continent.  Then there's the difficult 3k mile/5k km gap between Antarctica and Africa.

@Giovanni " so the question is where one would put the difference between an orbit and...something different. "  That question has been answered, at least in one way, with the Kármán line.  https://en.wikipedia.org/wiki/Kármán_line  At a given altitude we can compute the orbital velocity.  At a given altitude we can compute stall speed.  The line between flying/gliding/soaring and orbiting is when v-stall equals orbital velocity.

@MacGuffin If the orbit is elliptical enough you can reach a perigee below the Kármán line and continue orbiting the Earth. But sailplanes are far from orbital velocity anyway.

@Giovanni  In order to orbit below the Kármán line one must mix two definitions of the term.  One definition is to pick a number for legal reasons, another is to calculate the Kármán line for the given craft and conditions.  If a craft is in orbit then it can't cross below the calculated Kármán line, it can only cross the legally defined Kármán line.  It's been said that with enough thrust a pig can fly, I'll add that with enough velocity a sailplane can reach orbit.

@MacGuffin The Kármán line just defines the altitude at which you have to achieve orbital velocity (~4.8 mi/s) in order to fly with enough aerodynamic lift not to stall. It doesn't mandate what altitude is low enough to be unable to keep orbiting. That is rather something that the atmospheric pressure versus size and mass of the orbiter does. If the orbiter goes low enough, it would be unable to keep orbiting due to atmospheric drag, or would even burn due to atmospheric friction. But some consider an orbit if you just circumnavigate the Earth without propulsion, I guess that's what Ian meant

@Giovanni  " It doesn't mandate what altitude is low enough to be unable to keep orbiting. "  Below the Kármán line a craft is experiencing aerodynamic forces that dominate over ballistic motion of an orbit.  The craft will have to fly through the air in this case like any aircraft would, as in it's no longer in orbit at this point.  It can return to orbit should it go above it's Kármán line but below that line it cannot technically orbit.  There is a definition of an object or vehicle in orbit, and that is the Kármán line.  A sailplane is not in orbit unless it is above the Kármán line.

@MacGuffin All satellites that orbit within the thermosphere and lower exosphere experience atmospheric drag, even at say 400 miles (640 km) altitude. At lower altitudes the air just becomes thicker. So it depends on where the air is thick enough for you to consider it flying like any aircraft. This border is not the Kármán line. The current altitude record for the highest non-rocketplane in leveled flight is ~97,000 ft (29.5 km) by the AeroVironment Helios.

@Giovanni  I know what the Kármán line is.  As originally defined below that line a craft is in flight, above it a craft is in orbit.  There's another way to define it, a legal way, we pick a number that is convenient to define entering outer space.  This can be 50 miles, 60 miles, 100 kilometers, 80 kilometers, or whatever approximates the first definition for matters of things like violating a nation's airspace.  A craft can only orbit below the Kármán line by mixing the two definitions.  I'm repeating myself as I don't know how else to explain it in 600 characters or less.

@MacGuffin The round values you mention aren't really in use as political borders. The Space Shuttle flew below 50 mi (80 km) above foreign countries without consequences. On the other hand, some equatorial countries define their territory up to the geostationary orbit (~ 36,000 km / 22,000 mi).

To complement Peter's answer, let me mention that the farthest a glider has ever flown is 2256km (actual distance record is just beyond 3000km, but that was not in a straight line, but going back and forth). This is 1/17 of the proposed circumnavigation distance. And these flights have been done in exceptional days, in an exceptional area (Argentinean Andes). It is not easy to find the combination of well-oriented mountains and good enough wind and atmospheric conditions. There is simply not even a remote chance of chaining reasonable enough conditions to fly accross the globe.
Thermaling could help to fly the between two good mountain areas theoretically. In practice, excellent conditions are not prevalent, and it is common for thermal days to be irregular. The idea of leaving one mountain chain and have the thermals available to reach the next mountain chain, and to arrive and find the wind blowing in the needed direction; and then repeat such feat tens of times, over days, is way beyond the reach of imagination.
Not to mention human endurance. There was a time where endurance gliding was a thing. The single seater record sits at 56 hours; twin-seater, at 71. These kind of flights are not done anymore and possible records not recognized, because it was really dangerous. To fly 40000km at 200km/h requires 200hours, so almost three times the most anybody has flown in a glider. And this is assuming a kind of speed that is only achieved in selected mountains on selected days.

I think your endurance claims assume there is no autopilot - the endurance record is from 1961.

@AnonymousPhysicist Gliders without propulsion don't have autopilots anyway, do they?

@AnonymousPhysicist: an autopilot for a glider would be orders of magnitude above the pay grade of your usual autopilot that keeps the wings level and points the nose in certain directions. And no glider has ever been built that could possibly carry the crazy big batteries you would need to run the instruments for weeks; and that without the additional requirement of the several electric motors an autopilot needs.

It seems to me that a hypothetical endurance glider with an autopilot would have no batteries.  It would be solar powered.  The pilot would operate the glider at night.  The pilot could also take control when it was necessary to climb.

@AnonymousPhysicist: do you fly gliders? To keep a glider in the air, in particular if the goal is to cover significant distance, the pilot is working and making decisions very close to 100% of the time. And, there are cloudy days, too.

Solar cells work on cloudy days.

Even if you had solar panels, they would need to be connected to batteries to provide power at night. But the presence or absence of batteries isn't really the issue here. Outside of occasional research projects, there aren't existent autopilots that can even try to keep a glider in the air, and certainly not any that can operate independently with the pilot asleep.

@ZachLipton The obvious solution is to have two pilots who can take shifts. Possibly two pilots who suffer from dwarfism, to keep the pilot weight as low as possible.

Weight, surprisingly, is not such a big issue in gliders...

No. Both night and oceans would be major obstacles.
Night time can be survived by parking the glider in ridge lift but it requires a steady wind during the night. This is possible and has been demonstrated decades ago, but is only possible with the right weather conditions. To have those lined up between thermal-powered daytime flying is rather unlikely.
Oceans are too homogenous to create thermals. You will notice that when crossing a large body of water in a glider: Over water the air is calm while over land it is easy to find thermals. Why else would albatrosses resort to dynamic soaring if the much less stressful thermalling would be possible? Dynamic soaring works much less well for human-sized gliders because of the limited height of the wind shear layer above the ocean's surface. Besides, it is beyond normal human capability to fly intermittently at 3g over many hours without sleep.
If you start high enough, a large lake can be crossed since the lack of thermals means you also will not hit downdrafts and your glide ratio will be easy to predict. But no thermal will carry you high enough to hop between Pacific islands or from Scotland to Iceland, which is a nearly 1000 km overwater trip.
For a true circumnavigation you would need to touch the southern hemisphere and there are too many long stretches of water to plan any feasible course. Only flying across Siberia and on to Alaska, Canada and Greenland does not make it a circumnavigation. Also, I very much doubt you will find thermals over large bodies of ice and snow, too.

Comments are not for extended discussion; this conversation has been moved to chat.

Stick thermals do not depend on weather or geography, and are an extremely reliable source of lift.


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Does hypoxia take effect more quickly if an aircraft is flying at a greater height?
In case of depressurization, will the effects of hypoxia be more rapid the greater height the aircraft flies?

Related: How long does it take for pilots to become unconscious after an uncontrolled depressurization at FL 340?

Also Time of Useful Consciousness (TUC) at SKYbrary

I'm not sure what the root confusion might be...  If you know what hypoxia actually is, doesn't the question answer itself?

I did a chamber ride to 25000 once as part of a skydiver high alt cert course.  It was just to experience and become aware of your onset symptoms. I lasted about 4ish min, then tunnel vision set in (peripheral vision greyed out), then at maybe about 5 min the euphoria came on, in a gush that surprised me. I reconnected my mask and the symptoms vanished after the first or second breath. Buddy who was a smoker was well into euphoria stage and cackling like he was on laughing gas by 3-4 minutes.  Weirdly you are drawing shapes and you know they are lopsided, but you think they are fine like that.

The higher the altitude at which the plane depressurizes, the quicker the onset of hypoxia- and the less time the pilot has to get his or her oxygen mask on before passing out.

The "partial pressure" is too low.  Problem is, you don't know it when just sitting there.  You can't tell unless you are under exertion.  You don't realize there is even a problem until you enter the euphoria stage, and then it's too late unless there is someone else to intervene.

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What is the maximum Mach number that can be achieved at the exit of CD nozzle?
I understand that you can get Mach 1 at the throat of a convergent-divergent (CD) nozzle and after that you can have two roots, one is for subsonic and other is for supersonic.
Considering the supersonic curve, if you still manage to pull down the back pressure then you would get an underexpanded flow (which obviously we all want to avoid). So that sweet optimum low back pressure is desirable before it gets into underexpanded region for the maximum velocity at the exit of the nozzle.
I would like to know is there a way that we can increase the exit velocity even further (without getting into the underexpanded zone). Does maximum exit velocity not vary if you compare a small CD nozzle and a big one?

In rocket terms, maximum exhaust velocity is mainly a function of the fuel/oxidizer combination and chamber pressure, and produces a figure of merit called "specific impulse."  This is directly proportional to exhaust velocity.
Note that nozzle input pressure (chamber pressure, in a rocket) is a major contributor here.  A Merlin engine (used in the Falcon 9) and an F-1 (used in the Saturn V booster stage from the Apollo program) use the same fuel (highly refined kerosene) and oxidizer (liquid oxygen) but have significantly different specific impulse, because the Merlin operates at much higher chamber pressure.
Generally, overexpansion is the thing you want to avoid in nozzle design; this causes pressure zones and standing shock waves inside the nozzle bell that can damage the structure and result in an engine failure.  Underexpansion just means the exhaust has some level of unextracted energy when it leaves the nozzle.  Underexpansion is very common in real rockets; Falcon 9 launches exhibit it quite visibly (in the form of the very wide exhaust plume) for the last minute or so of the first stage burn; this is the case with all orbital rockets launched from the ground, because the engines must be designed to work well at sea level pressure.
Significant overexpansion is seldom seen in practice -- but a recent example was when SpaceX Starship 20 test fired the one Vacuum Raptor engine that was installed at that time, just a few days ago, with the vehicle clamped down on the test stand at sea level.  More commonly, engines are designed with a compromise expansion value allowing safe operation at sea level and less underexpansion at maximum operating altitude (typically around 80-100 km -- effectively vacuum -- for a booster).

Almost all booster engines are overexpanded on the pad. You might want to qualify your last paragraph. See 2nd paragraph in this paper. https://ntrs.nasa.gov/api/citations/20070031864/downloads/20070031864.pdf

given the focus on rockets I feel this should be moved to space.SE

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What are the design considerations for the ailerons of the Pilatus PC7 Mk II?
The ailerons trailing edge is not in line with the adjacent inboard wing TE, jutting out by few cm:

What are the design considerations?

That's an aileron trim tab and/or servo tab. I looks to me a bit of chord was added after development to improve the roll rate while leaving the inboard fixed TE, under which is a split fowler flap, alone.


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How are individual spool optimal speeds determined in 3-spool engines like Trent 700/800?
I can see that in a 3-spool engine, the low pressure turbine and the fan have the lowest speed; The intermediate pressure turbine and the intermediate pressure compressor have another speed, and the last spool, which is the high pressure turbine connected with the high pressure compressor, has the highest speed.
As far as I know, Rolls-Royce implemented this design on Trent 700/800 engines to insure every part of jet engine works at its optimal speed.
My question is: How can they find the best speed for each spool? What tools are used? Or are tests conducted to figure out exactly the optimal speed per spool?

The most determinant factor is air pressure varies along the compressor stages, and along the turbine stages. This requires adjusting the air duct cross-section, and in turn the blade size, which implies changing the blades angular velocities.

Brief engine description
Taking only the compressor into account (the reasoning is similar for the turbine): It is composed of stages, each stage is made of a rotating disk with blades (rotor), and stationary vanes (stator). The rotor role is to accelerate air, the stator role is to convert the excess of air velocity created by the blades into pressure.

(Trent engine, source: Danish Anwar's work at Behance.net)
Hence, as air progresses along the compressor ducting, pressure is progressively increased while velocity is kept more or less constant. The duct itself is annular with inner and outer walls. Blades and vanes are located between walls, vanes are fixed to the outer wall, and blades are fixed to solid rotating disks forming the inner wall.
General annular duct design
At each stage, air velocity relatively to the rotor blade is the vector sum of the flow velocity along the engine axis and the blade tangential velocity, the latter varying along the blade length. It means the total air velocity relative to a blade depends on the angular velocity of the rotor, the outer diameter of the duct and the blades length, which can all vary between stages. There is a first choice to be made: What should be the duct inner and outer diameters for each stage? But this can't be answered directly.
As air moves from one stage to the next, its mass doesn't change but its volume decreases as its pressure increases. The design of the next stage must take this into account: The cross-section of the annular ducting must be smaller. This means changing the geometry of the blades (and vanes of the same stage), either the diameter of the rotor disk, or the length of the rotor blades, or both. This is another choice, based on air velocity where it is the largest, at the blades tip.
If the next stage is on the same spool, the angular velocity is the same but velocity of air at the blades tip is different due to the different outer diameter. This means some compromise is required to select the angular velocity and/or the diameters.
Blade efficiency
Engineers have a general design constraint: Compressor efficiency. Pressure increase is provided by vanes according to air acceleration, efficiency is first a matter of efficiency in air acceleration by the blades. This is achieved by using a high flow velocity at the blade tip, a fraction of the speed of sound (speed of sound for the fan).
The compressor must deliver the air flow required by the combustion chamber. The more powerful the engine, the larger this flow. This can be achieved by:

A larger air duct cross-section, which leads to a larger duct diameter and a larger engine, or
A larger angular velocity which leads to more noise and possibly a lower efficiency.

This defines the outer diameter of the duct for a given stage. As the cross section is defined by the pressure at this location, the inner diameter is also dependent on the outer diameter.
There is a compromise to be accepted: Velocity varies along the blade length. The same cross-section can be obtained with a shorter blade if the outer diameter is larger, this limits the velocity variation along the blade length. A smaller variation means mean velocity is closer to the optimum. But a shorter blade requires a larger chord (width), decreasing the aspect ratio, which in turn decreases the efficiency, so another compromise is to be decided when selecting the outer diameter of a stage.
Mass and other mechanical considerations
Engineer have more constraints, including minimal mass, minimal deformation, and minimal lag when changing thrust, which influence the blade size, hence the duct shape and the angular velocity for each stage.
However having a different angular velocity for each stage would add mechanical complexity, mass and weak points. The current choice is to limit the number of different velocities to two or three, giving engines with two or three spools. Rolls-Royce designed a 3-spool engine in the late 60s (RB.203), and maintained this choice until today with the Trent family.
Spool velocity
For a given spool, choosing its angular velocity is a matter of averaging the angular velocities or all the stages on the same spool. This can be a challenge, especially if the engine is a turbofan: The fan is used both for the cold flow (bypass air) and the core flow, it cannot be efficient for both flows unless using a dedicated spool for the fan.
A reasoning similar to the previous one can be applied to the turbine stages. Indeed the number of angular velocities in the turbine must equal the number of angular velocities in the compressor. For each spool the turbine section role is to spin the compressor/fan section on the same spool.
The angular velocity of a spool is also a compromise between the angular velocity required for the compressor and the one required for the turbine. In the process of optimization, the tip velocities and the duct diameters can be adjusted for the optimal match.
Stall prevention
In order for the compressor to work, the air flow must be continuous. The engine is optimized for a working point (take off or cruise) and the different spools are optimized to deliver this homogeneous flow in spite of the different angular velocities.
When thrust demand is changes a new fuel rate is set in the combustion chamber, this introduces a new regime for the engine, spool velocities are changed. This can lead to breaking the flow homogeneity between spools, leading to blade stall, especially for a 2-spool engine, as the design point may not be as optimal.
Stall prevention greatly relies on the introduction of another parameter, the vane angle of attack of the first stages. For some Trent engines, it consists in the first four stages of the intermediate spool being equipped with variable (adjustable AoA) vanes.
Environmental constraints
Constraints like noise and pollutants rejection minimization greatly influence the velocity the fan and its size/mass. Using three spools, with one dedicated to the fan, might be a great advantage, each spool being better optimized, meaning less fuel required to spin it, hence less pollution.
Conclusion
In conclusion, a spool angular velocity is a compromise between between multiple factors, in order to find the best efficiency:

Best match between compressor and turbine sections
Best match between duct diameters for the stages on a given spool.
Best mass/velocity choice for the blades
Best homogeneity in the flow to prevent compressor stall.

This is first a matter of computation: Duct area and air velocity at the tip of a blade are straightforward. Finding compromises between stages given a number of spools is also relatively easy. Then tests on the engine prototype are conducted to check the validity of the choices at varying thrust demands, adjustments are done.
See also: : Why does the Trent XWB have intermediate compressor stages?


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Do some states allow a substitute for runway base course material, in lieu of the FAA P-209?
I would like to know if, for a runway, some State can allow a substitute for the base course in lieu of the FAA crushed aggregate base course (P-209) defined in FAA AC 150/5370-10.
E.g. Can Nevada allow a base course defined in Nevada DOT Spec?

Please explain what you're referring to when you use the term "base course".  Also, what is an "FAA P 209", please?  Typically, states don't regulate the things that the FAA regulates, i.e. most things in aviation.

Is your question strictly related to airports that need to be certified under FAA Part 139?


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14 CFR Part 101 -- Moored Balloons, Kites
I am hoping someone can hep me interpret the Regulations regarding moored balloons. I want to fly a 5-1/2 ft lighted helium balloon at night at 500 feet AGL.
I have made exhaustive searches online, contacted and gotten a response from the FAA by email, and spoke with the FSDO responsible for my area, but I still don't feel like I have a resolution.

I feel like since the balloon is less than 6' and none of the other title 14 101 Subpart A 101.1 Applicability criteria apply, then my balloon is not subject to any regulations in Part 101.

The FSDO kept telling me all they can do is refer me to the regulations, they couldn't talk openly about my situation.

I got the impression the FDSO didn't know, I asked if a 8" helium balloon required an automatic rapid deflation device and they also said "all I can do is refer you to the regulations".

Because your balloon is less that 6' diameter, you are correct in presuming the subpart 101.1 you are asking about does not apply. However, other parts may, and the FSDO isn't going to research this for you or give you blanket permission...

You can also try law.se for how to interpret this specific section of the law, and know the possible jurisprudence, there are also Part 101 comments. Local regulations may apply (e.g. Dallas) and may add constraints not in Part 101.


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Was there ever a supersonic flight in a non-pressurized plane, and what is/would it be like for the pilot?
If you went supersonic in a plane that has no pressurized cockpit, would you be completely deaf (because of flying faster-than-sound)? Would it be dangerous to the pilot even if he/she was in an enclosed cockpit, and if so, is this the reason why we still don't have supersonic maglev trains? Did anyone ever attempt or conduct a supersonic flight in a non-pressurized plane?

It would be as safe as any other non pressurised plane at that altitude

@Abdullah I didn't specify an altitude. Would one be deaf when supersonic?

No, one would not be deaf, because the air in the cabin is moving with you. On a related note, if you jump from a Boeing 737MAX and hit another Boeing 737MAX flying the other way, you would break the windshield, then break the sound barrier.

@Abdullah But the cabin isn't pressurized, and some general aviation aircraft have tiny open holes/windows where you can stretch your handpalm out. If a plane with such windows flew supersonic, would the air still move with you so to not be deaf(er)?

May be a duplicate. Answered here: https://aviation.stackexchange.com/questions/17056/is-it-quiet-in-the-cockpit-when-flying-faster-than-the-speed-of-sound

@JohnK I knew what is told there, I just wondered whether this applies to non-pressurized planes too (and it doesn't as per Arkhem's comments).

Very interesting comment by @Abdullah, but reversing subsonic speed in the opposite direction would only break the unfortunate involved.  Important, because this is about relative motion.  Also of note:  how much would Bernoulli lower the pressure in a unsealed cabin.  Lastly, how much more thrust is required to go supersonic at a survivably low non-pressurized altitude?  Probably why the B-1 folded its wings.

@RobertDiGiovanni Bernoulli would not necessarily lower the pressure much. It actually compares the stagnation pressure with static pressure.

@Abdullah Thanks for the answer in travel.SE. Unfortunately I can't accept it because I don't have an account there. For those who're curious: yes, there are meals in current passenger flights and you can put your mask off then.

When you're in the cockpit, the sound you are hearing is vibrations of the skin and canopy/windows induced by the outside flow, transmitted through the air within the cabin to your ears.  A very loud hiss/roar (even in a normal jet, if you speed up to near Vmo, say .85 M, the noise level in the flight deck gets unpleasantly loud.  Go faster, and it's the same hiss/roar but the more energy being imparted means the sound level transmitted to your ears goes up).
When Chuck Yeager was going supersonic in the X-1, he could only tell he was supersonic was by the Mach Meter, not by the noise level.
If the cockpit was unpressurized, this doesn't really change because it's still just noise transmitted from vibrations of the skins/windows transmitted through air that is stationary relative to objects it's in contact with, so you're still just hearing a hiss/roar of the outer airstream passing by, supersonic flow or not. Just way louder, I'm sure, because of less sealing and insulation.
Any other sounds originating outside you would also hear to the extent that the sound waves from that sound impacted that skin and caused a change in the sound character being transmitted through the stationary air in the cockpit.  If the wave was intense enough, you should hear a thump sound of some sort, whatever was the result the pressure wave of the sound origin changing the transmitted energy passing through the still air in the cockpit at the instant it reaches the fuselage.
The key part is that the air within the cockpit is not moving with any speed, so it will transmit any vibration from the skins, windows, engines, pumps, etc to your ears whether you are supersonic or not.

And if your cockpit wasn't enclosed, you'd sit in open air instead, you wouldn't hear anything as Arkhem stated, unless there was a sound in front of you, right?

@Giovanni: if you were sitting in the open air with it moving past you at > 1235 km/h, I'd say you would have more pressing things to worry about than the sounds you hear. But yeah, I guess you wouldn't hear sounds generated behind you, unless they are transmitted through the fuselage...

@rob74 What things to worry about exactly? I'm assuming here the pilot flies below the Death Zone altitude (< 26,000 ft), and even above that altitude you don't necessarily need a pressurized or enclosed cockpit, just artificial oxygen supply.

@rob74 The 1235 kmh number is just for 20°C. In the upper troposphere sound is considerably slower. Since your planet's mean surface temperature is 14°C, I don't get why the speed of sound is usually specified at 20°C.

@Giovanni A lot of scientists are past 50 years of age, and their fingers get too stiff to take good notes when the room is down to 14C.   Yes, I know, that's what post-grads are for, and they're generally in their 20s.  It's mostly a joke, anyway...

I refer you to Thrust SSC, a supersonic car.
The sonic boom is left behind you as you are by definition going faster than the sound. It’s the people around you who are damaged by it. Hence why no supersonic flight over land - and news articles when it does happen for security reasons - and no supersonic vehicles in general.

But one is deaf when driving or flying supersonic in a non-pressurized cabin?

Do you mean “are you deafened” (the noise damages your ears) or “is all sound left behind you so that there is nothing for your ears to hear?” If there were sound being generated in front of you then you would pass through the waves and hear it. However you would not hear sound behind you.  A greatly simplified example: If you are riding your bicycle next to someone in a car at speed and they throw some water in front of you from their window you would get wet as you would catch up with it. If they throw from behind you you would not get wet as you are moving forward faster than the water is.

I mean your 2nd question, all the sound is left behind you. So your answer is yes, you'd be deaf because the sound (the water in your example) would remain behind you?

@Giovanni I wouldn’t use the word ‘deaf’ as that implies you are incapable of hearing because your body is damaged or defective. You aren’t deaf, there’s just nothing to hear unless you pass through the sound wave of something in front of you. I suspect it would sound odd through doppler shift but I defer to physicists for that.

I see, thank you.

I didn't find a better word for hearing nothing, so I belet it on 'deaf', but I clarified it in my question summary.

I was thinking about what if a sound occurs indeed in front of you, e.g. a thunder one mile before your plane's nose? Would you hear that? I think you would hear it. Just like if you travelled faster-than-light, you'd see nothing behind you but you'd still be able to see light sources in front of your spacecraft, right?

I believe the Bell XS-1 was unpressurized.  As John K points out, pressurization will not have an effect on sound levels in the cockpit.
In regards to your comments about supersonic trains, the energy consumption at low levels would be prohibitive in order to overcome exterior parasite drag and there are significant problems keeping and maintaining a vehicle on a track at those speeds.  Attempts such as the Hyperloop in order to combat parasite drag by operating the vehicle inside of an evacuated tunnel are being experimented with, but in my opinion do not show a whole lot of promise, as the entire system would be cost prohibitive to operate, not to mention riddled with design difficulties.  It has nothing to do with a pressurized cabin versus an unpressurized cabin.

I also wonder what is it like to drive or fly supersonic while you're in open air (such as a cabriolet and its counterpart among airplanes).


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Are flights being shown as "hijacked" on airport departure screens in case they are?
In the movie The Net a flight appears as "hijacked" on a departure/arrival screen. Would these screens really show an airliner as hijacked when they are, and did this ever happen?

In the movie it's the LAX (Los Angeles) airport if we wanna determine it in its specific case.

I seriously doubt they would show that, but questions about departure screens in airport terminal buildings are off-topic here anyway.

@Bianfable I wouldn't say that. It belongs to airport operations so it fits both aviation and travel.

Questions about whether something is possible or not are rarely useful, unless there is a plausible rationale given for why it might not be possible, and addressing that rationale can provide some meaningful insight.  There is no reason to believe that computer display couldn't be programmed to say any words you want it to say, therefore I'm voting to close and DVing.

@MichaelHall The question was whether it is being shown on-screen if an airliner is actually hijacked.

@Giovanni, you asked "Can a flight be shown..." and "do these screens have the possibility..."  If you want to ask whether it is a good idea, or whether airport policy would be to display this information, then please rephrase your question to say that and I would be happy to remove my DV and close vote.

@MichaelHall Alright, I rephrased it.

In the airport, I would guess (without any knowledge) "SEE AGENT" would come up before "Hijacked".

I don't have any experience with the software used (at LAX or anywhere else) but here are my thoughts.
In theory anything can be shown if the display is connected to a general-purpose computer, but I doubt the software displaying flights and times has "hijacked" as a selectable status, or that it has a free-text entry option.
The hack would, in all likelihood, have to be an edit to the compiled binary/executable of the program that is shown on the screen, similar to the story about the "Thank you for playing Wing Commander" error message. This could change all instances of "Delayed" to show "Hijacked" instead (though there are more letters in "Hijacked" so this may not even be possible). The string "Delayed" could also be stored outside the binary in a manner similar to the .lproj/.strings system Apple uses, which would make editing it easier. But in both cases the change would be seen for all delayed flights rather than just one.
Or it might be that the flight information is stored in a database separately before being passed to the binary which displays it on the screen; if "flight status" is stored as a plaintext string instead of a numeric code (to be decoded within the binary) it would be very possible to change the status for one flight in particular to be anything you want. Personally I doubt this is the case because the system needs to get information automatically from some central location and is continually updated by feeds from the airlines. But the hackers may have disabled automatic updates as well.

Yes, the hackers (Pretoreans) were really powerful in the film. The screen changed permanently, it was just to cause chaos. I'm sure multiple planes were labeled "hijacked" on the chaotically changing screen.

Anything can be shown on a digital screen, but there is no way any airport would do that. What would be the purpose?

To tell passengers the truth. But in the movie, it was actually a hack on the airport computers.

@Giovanni imagine you've gone to the airport to pick up your loved one and check their flight status and see that... it's a lot more humane to tell you the bad news in person


text:
Why do airplanes have complex air conditioning systems?
I'm wondering why do planes have complex air conditioning systems with several coolers, compressors and turbines. If these devices are powered by bleed air from the engines, why isn't the bleed air simply cooled and depressurized with valves? Is it due to easier pressure regulations as the pressure of the bleed air surely changes depending on the amount of current thrust from the engines, altitude of the plane etc. or is there some other reason?

It's because for the Air Cycle process, to chill ambient air without refrigerant, to work, you have to take really hot high pressure air and and remove most of the heat while keeping the pressure up, so that when you let the pressure drop at the end of the process it ends up way below ambient temperature.
So you take air off the engine compressor at some pressure like 80 psi and, say, 400F, pass it through a radiator (precooler) to cool it down, pass it though another compressor (the Air Cycle Machine's compressor) to boost the pressure, and therefore the temperature, again, run that through more air to air radiators or heat exchangers to cool that down, then, when you have the pressure still really high but have removed most of the heat, you let it expand and drop the pressure closer to ambient.  While doing that, you pass it by a turbine, which drives the compressor.  (An Air Cycle Machine is more or less a big turbocharger like device, pipes, radiators, and valves).
You now end up with air that entered the engine at 70 deg F coming out the Air Cycle Machine at just above freezing when it's properly regulated.  That air is mixed with raw hot bleed downstream to create the final inflow to the cabin.
An ACM can in theory put out extremely cold sub zero air if all the bleed flow available is run through it.  That is to be avoided because ice formation on the expansion turbine and inlet nozzles becomes a problem and this is one of the main factors in damaging them.  The other headache is dealing with all the condensate water, which causes turbine erosion problems.

Also, jet engine compressors (unlike turbines) are inevitable potentially unstable machines, and the bleed air needs to be controlled to prevent compressor surges and stalls. Just regulating the engine bleed with a simple "valve" (controlled by a thermostat in the passenger cabin, or whatever "simple" system you might be thinking of) would not be a recipe for reliable engine performance.


text:
What is a second oversize Hi-Lok fastener?
Many structural repair guidelines specify to use a 2nd oversize fastener e.g.3/16 second oversize Hi-Lok, when replacing an existing fastener in an assembly.
What does "second oversize" mean in "3/16 second oversize Hi-Lok?

When a hole has been damaged, its diameter is increased to get a new clean hole and recover the structural integrity. This requires the use of an oversize fastener (alternatively a sleeve).
3/16″ is a reference to standard diameter and "second" means the second increased diameter. From Jet-Tek catalog, the diameter is increased by steps of 1/64″:

The standard diameter is ⌀ = 3/16″ (or 6/32″ or 12/64″).

The 1st oversize (+1/64″) is ⌀ = 13/64″.

The 2nd oversize (+1/32″) is ⌀ = 7/32″ (14/64″).

The oversize diameter is always between two nominal diameters. The part number is often common to the standard and oversize fasteners, with a unique character representing the oversize.

+1 Correct, except that I don’t think the 3/16” can be referring to the increase in diameter - that would be huge, representing more than the largest step between standard sizes. More likely “3/16 second oversize” refers to a 3/16” nominal shank + two oversize steps, i.e. .21875” shank.

You're right, corrected. Thanks.

How do you "damage a hole"? :-)


text:
What are fuselage Inserts?
From weight & balance control and loading manual:

Body Station (B.S.) is defined as a manufacturing location on the airplane. For first of an airplane model, B.S. are continuous from the front to the aft of the airplane. For later versions that are either stretched (i.e. fuselage inserts added) or shrunk (i.e. fuselage sections removed), B.S. becomes discontinuous, for manufacturing reasons.

What are fuselage inserts?
Why are Body Stations discontinuous when they are added to the fuselage?

Unrelated to the question, but I can't help seeing your username as "skywaitress"...  Just in case you care about such weird little things.  ;)

Which weight & balance control and loading manual?  Are you referring to the FAA handbook from the USA?

what are fuselage inserts? and why is Body Station discontinuous when they are added to the fuselage?

Body stations ($\small \sf BS$ or $\small \sf STA$) are used to describe the position of the fuselage sections. They correspond to the distance of the fuselage frames from the STA reference point. The reference point varies among designers, for the Boeing 737 NG series it is a point 130.0 inches forward of the airplane nose, and a $\small \sf STA$ number corresponds initially to the distance from this point in inches.

See What is the coordinate system used in an Aircraft?.

The fuselage sections themselves are major aircraft zones, e.g. section $\small \sf 43$ on a 737 NG:

Source: Boeing 737 NG Maintenance Manual
Manufacturers design families of aircraft to suit different passengers or cargo configurations, e.g. the Boeing 737 NG family includes different fuselage lengths for models -600/-700/-800/-900. The fuselages are almost identical except a few sections are adjusted, a.g. section $\small \sf 43$  can be extended by inserting additional segments, the inserts of your question.
For example, the sections of the 737-600 fuselage have $\small \sf STA$ numbers corresponding to actual distances from the reference point. To create a  737-700 fuselage, section $\small \sf 43$ space between $\small \sf {STA \; 500}$ and $\small \sf {STA \; 520}$ (20" long) is extended with stations $\small \sf {STA \; 500A}$ through $\small \sf {STA \; 500D}$ (80" long). This adds 3 windows on each side of the cabin. The corresponding diagram for the 737-700:

Source: Boeing 737 NG Maintenance Manual
The numbering is still monotonic, but numbers beyond $\small \sf {STA \; 500}$ don't correspond anymore to distances from the STA reference point. Thus when checking W&B with loads specified using $\small \sf {STA}$ numbers a conversion table [station ⇄ arm] must be available.

Much obliged....


text:
How to calculcute variation and compass deviation in this CDMVT problem?
Usually I can solve CDMVT problems, but sometimes I feel I'm missing something important. For example:

I'm able to solve:
Given WCA -012, TH 125, MC 139, DEV 002E. What are TC, MH and CH?
T     V    M    D      C
125 + 2 = 127 - 002E = 125
————————W(+)——E(-)———————————->

TH=TC+/-WCA
MH=TH+/-VAR
MH=MC+/-WCA
CH=MH+/-DEV
125=TC-012 => TC = 137
MH=139-102 => MH = 127
CH =125

Using the same method, I cannot solve:
Given: TC 183, WCA +011, MH 198, CH 200
T     V      M     D     C
194 + 4  =  198 + 002  = 200
————-W(+) —-E(-)——————->

TH is easy 183+011=194
Variation looks like +4 so 4W, and deviation looks like +002.
But the correct results are variation is 004E (so minus 4)
and DEV is -002 not plus.

Please someone clear me up what I am missing I feel like is important and I miss something critical.

Abbreviations used:

TN/MN: True/Magnetic North

VAR: Magnetic variation, positive when MN comes after TN, clockwise, said otherwise when VAR is east.

TC/MC: True/Magnetic Course. The track actually flown

TH/MH: True/Magnetic Heading. The aircraft heading.

WCA: Wind correction angle, angle added to desired course to get the heading to be flown.

CH: Compass reading

DEV: Compass error (deviation). Quantity added to compass reading to get the actual magnetic heading.

Thus we get:

MH/MC = TH/TC + VAR
MH = CH + DEV
TH/MH = TC/MC + WCA

What are TC, MH and CH
given: MC = 139, TH = 125, DEV = 2, WCA = -12 (given are between brackets)
          C  +  D --> M  -  V --> T
Course              [139]   2    137
WCA                             [-12]
Heading  125   [2]   127    2   [125]

Answer: TC = 137, MH = 127, CH = 125
Because:

TC = TH - WCA = 125 - (-12) = 137
VAR = MC - TC = 139 - 137 = 2 (meaning MN is east of TN)
MH = TH + VAR = 125 + 2 = 127
CH = MH - DEV = 127 - 2 = 125

What are TH, VAR and DEV
given: TC = 183, WCA = 11, MH = 198, CH = 200
          T  +  V --> M  -  D --> C
Course               187    4   [183]
+ WCA                            [11]
Heading [200]  -2   [198]   4    194

Answer: TH = 194, VAR = 4 (east), DEV = -2
Because:

MC = MH - WCA = 198 - 11 = 187
VAR = MC - TC  = 187 - 183 = 4 (so MN is east of TN)
TH = MH - VAR = 198 - 4 = 194
DEV = MH - CH = 198 - 200 = -2

I used the standard definition of the magnetic variation, positive when MN is eastward. You seem to prefer positive when MN is westward, just invert MH/MC = TH/TC + VAR, that is use: TH/TC = MH/MC + VAR and you'll get the same result.


text:
How do changes in airspeed in a level, constant bank uncoordinated turn affect coordination?
I'm hurting my brain trying to understand how the PHAK talks about correcting slips and skids. I think part of my confusion comes from the fact that the formulas for rate and radius of turn assume a coordinated turn.
To paraphrase, the PHAK says that to correct a slipping turn, you can decrease bank angle, increase the Rate of Turn, or a combination of both.
I think some ways to increase Rate of Turn would be too input more inside rudder and/or apply more back stick pressure. Would another be increasing airspeed if bank angle is held constant (assuming a level turn)?
This goes against those formulas I mentioned, but like I said, I think know that they only apply to coordinated turns. Thanks

As Robert DiGiovanni said, normally you would use any combination of aileron/rudder to change your rate of turn. Yes, changes in pitch (and hence changes in total lift produced) also has an effect because it is a component of lift that pushes us in to the turn but in practical flying we would reserve pitch/power changes to correct for ALT/VS and aileron/rudder for turning.

Normally one corrects for a slipping turn with less aileron or more rudder (watch the ball).
Increasing the rate of turn could be accomplished by slowing down.

in the ROT formula speed is in the denominator

Since the plane is banked, more elevator input to pitch for lower speed also creates more inward (centripetal force) relative to the forward force (thrust) vector.

but slowing in a turn can be dangerous

Best to keep it simple.  Coordinate rudder and aileron.  Use power to control climb and descent.  Yes, it works in turns too.

Changing power setting in a single will also affect its rolling tendency, so one could theoretically "correct" a slipping right hand turn by adding power, but again, the go-to method of coordinating is $rudder$.


text:
Is your instrument rating tied to category and class?
Reading through 14 CFR Part 61, the Instrument Rating Airplane ACS, and pilot forums, I seem to get conflicting answers to this question.
In the Instrument ACS, it says, "if an applicant holds both single-engine and multiengine class ratings on a pilot certificate and takes the practical test in a single-engine airplane, the certificate issued must bear the limitation 'multiengine limited to VFR only'".
Let's say I have a private ASEL with instrument and I go to get my ASES add on. Do instrument privileges transfer to when I'm flying a seaplane?

Yes.  If you take an ASES addon, you will have instrument privileges for that class aircraft.
The instrument rating is for INSTRUMENT AIRPLANE which is category specific and not class specific.
The reason why there is a MULTIENGINE LIMITED TO VFR ONLY limitation is due to the applicant not showing the examiner they are proficient in multi-engine one engine inoperative operations and approach.  The instrument ACS lists the required tasks needed on a multi-engine addon with instrument privileges.

AOO VII, Task B: OEI during Straight and Level Flight and Turns
AOO VII, Task C: Instrument Approach and Landing with an Inoperative Engine

There are two ways to get the VFR ONLY limitation on AMEL.

Pilot has ASEL and IA but does not complete the two above tasks during the AMEL practical test.
Pilot has both ASEL and AMEL and takes IA practical test in single engine airplane.

So why am I limited to  private & VFR-only for SEL? All my checkrides (and Military Equivalence) for everything after high school flying when I got my Private have been in MEL aircraft, but what SEL-flavor tasks are there that don't get checked during the ride in a multi?  By the logic in this answer, shouldn't my instrument rating apply to my SEL also?

@RalphJ I've never seen a limitation of "SINGLE ENGINE LIMITED TO VFR ONLY" on a certificate.  Do you have one?  I'd love to see it.  All instrument ACS tasks can be performed in a multiengine airplane.  If you pass in a multi, you get privileges for your single aircraft too.  This isn't true for the private or commercial ACS.  There are specific tasks there that must be done in either a single or multi-engine airplane.


text:
Is it possible to hear a sonic boom when the aircraft is exactly at Mach one?
Will an observer on the ground hear a sonic boom if a plane passes overhead at exactly the speed of sound? That is, the plane does not cross the sound barrier. The plane just hits Mach one - and continues to travel at that exact speed.
This is a bit of a problem because the speed of sound changes with altitude, so we assume it remains at the exact same altitude. I imagine there are other caveats also!
I would guess that there is a sonic boom, but I'm not sure.

Another caveat people often miss: if the plane flies at Mach 0.999, but the airflow over the wing is faster than the incident flow... is the plane technically supersonic?

@AEhere But for the sound energy to concentrate in a shock wave, you need to consider the environment between the plane and the observer. Things that happen very close to the plane are not interesting. Actually, all jet planes would be technically supersonic by this consideration as their fan is supersonic :-)

Yes but the speed of sound for that specific environment: temperature, humidity, density, etc

@AEhere When flow in some areas is supersonic and in other areas is subsonic it is called transsonic

Not only does it produce a sonic boom at Mach 1, it will be the loudest boom that that particular aircraft can produce.  Image source
In the picture, 4 is the shock wave. The one at M1 is vertical, and has all the concentrated pressure differential in it. An aircraft flying at supersonic speeds forms two shock cones, one at the nose and one at the tail, with a pressure distribution in between that spreads the pressure differential out.

But the total power contained in the shockwave must surely be higher at higher speed? Power is force times velocity, so I would imagine that the power dissipated by the shockwave grows quadratically with speed.

@Sanchises "power is force times velocity" -- why should it be so also for the vibrations the plane induces on the surrounding environment? (I don't say it isn't, I just don't see the reason.)

@Sanchises the output of the sonic boom itself is a step input in pressure at M=1, a more gradual pressure increase to the same end value when speed is higher.

@Sanchises the catch is in the figure .2 you see many shock waves "merging"/"adding" in a single big one boom

@yo' The engines deliver a continuous amount of power (which is higher at higher velocity), which needs to be dissipated one way or another. While of course there are many ways to dissipate power, I would expect a large part of this power to be dissipated through the shockwave.

@Koyovis I do believe you, but I think the explanation is a bit lacking. I can see that the shock is spread out over a cone rather than a flat disk (so I suppose it scales with $\tan\alpha$ or something), but how does the intensity of the shockwave come in to play, if at all?

No, if standard atmospheric conditions apply.
Since speed of sound is proportional to temperature and temperature normally decreases with altitude, the speed of the aircraft at Mach 1 is subsonic in warmer, lower air. This means also that the Mach shock will diffuse and be audible on the ground either as a protracted rumble or not at all, depending on the aircraft's altitude.
If you are close by, the boom is a sharp, very loud double-bang which can shatter windows. If the airplane is flying a few thousand feet overhead, you hear a muted rumble (still very distinct, though). If its altitude is above 10.000 ft and you are still on the ground, it will silently pass. Except in an inversion, of course.
Note that I assumed that the listener is located below the flight path of the aircraft. If that would not be the case, the noise of the aircraft diving below the listener altitude would be substantial.

Dear downvoters, if you don't understand physics but feel that an answer you don't comprehend is wrong, please educate yourself before rashly downvoting.

Yes, this answer is more sensible than some of the others

I think the issue is that you and @Koyovis both have given correct answers, but they seemingly contradict one another.  I might suggest you provide a sense of scale at which the altitude difference between the aircraft and the observer is significant enough to produce the atmospheric effect you describe.  (perhaps in just a bit more detail than "depending on the altitude"...?) Because there is a pretty huge difference between a Mach 1.0 Concorde overhead at 35K' and an F-14 Tomcat 200' off the deck.

In other words, maybe instead of opening with a hard "no" that those with a short attention span might not get past, try "yes, but..."

@MichaelHall You're right, that short attention span. I have witnessed dozens of sonic booms but not even in a single case together with flight data. I've heard the sharp bangs and saw the airplane close by and heard the rumble and saw it flying higher up, but without data.

@MichaelHall I think you're right about the seeming contradiction. I'm upvoting this one, but not Koyovis's one because this one provides factual principles which back up the claim (the very good observation that density increasing will mean the speed of the shock wave will be below the speed of sound).

@KennSebesta It's actually the increased temperature which will make the shock wave diffuse. BTW, I upvoted Koyovis's answer for the observation that the shock will be most intense at Mach 1.

Yes a plane at M1.0 would make a sonic boom. And yes it would be heard on the ground.
@Peter Kämpf and others who say no are demonstratively incorrect.
On Chuck Yeager’s first supersonic flight he took the X-1 to Mach 1.06…at 43,000 feet.
Mach 1 at that altitude is approximately 660 mph so 1.06 would be 700 mph.
He was flying over the Mojave desert, approx 5000 feet elevation. The speed of sound at 5000 is approximately 748 mph. Well above the 700 the X-1 was going.
His sonic boom was clearly heard by NASA observers on the ground.

And you know the atmospheric conditions that day, don't you? And you know what was heard was a sharp bang, not a rumble? Impressive!

@constructionist, do you have any references to support the data that you're providing?

@ROIMaison Where should I begin?  https://www.space.com/16709-breaking-the-sound-barrier.html   https://www.washingtonpost.com/sf/brand-connect/lockheed-martin/quick-and-quiet/   https://www.wired.com/story/chuck-yeagers-1947-flight-inspired-supersonic-ambitions/   https://airandspace.si.edu/stories/editorial/breaking-sound-barrier-75th

@PeterKämpf   Play at around 4:30 for 10 seconds or so.  https://www.youtube.com/watch?v=i_rFAo358bU

@PeterKämpf Or here starting at about 2:20.   "two sharp blasts roll across the desert"   https://www.facebook.com/SmithsonianChannelAviationNation/videos/the-moment-chuck-yeager-broke-the-sound-barrier/573843757763244/

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What is the difference between dry operating weight and dry operating mass?
What is the difference between dry operating mass and dry operating weight in the context of rotorcraft?
What is the formula for converting mass to weight?

You should provide the context, as in aviation weight is often used where mass would be more accurate. w=m*9.81 at sea level and at the equator. Mass unit is the kilogram, weight unit is the newton.

First you weigh the object on the moon, then you divide that number by 0.167 and that is the mass. Getting the aircraft to the moon is the tough part.

On Earth, there is, for practical purposes, no difference.

Weight is a force.
Mass is a property of matter. Acceleration makes mass produce a force. Specifically, a mass in the gravitational field of Earth effects a weight force on its support. The relation between both is the gravitational acceleration g, which is standardized to 9.80665 m/s² (approximately 32.174 ft/s²), as in Weight = Mass · g.

The problem occurs due to lazy terminology. In aviation both normally mean the same thing: The reading you would get if you put the empty aircraft onto a large weighing device (scales, or whatever). This is normally given in kilograms (kg) or pounds (lb).
This trouble arises because both of these quantities are actually masses, the weights would be in Newtons (N) or pounds-force (lbf). To transform the mass into a weight you multiply by the local gravity, g, which is usually assumed to be 9.81m/s2.
A similar problem exists when at home: If you weigh yourself on standard bathroom scales you will normally see a value in kilograms (i.e: A mass), but the scales actually measure a force. This means if you use the same scales on another planet you will get a different value in kilograms, which is nonsensical. The same goes for aviation: Weight and mass are being used interchangeably, which is incorrect.

edit - And to specifically answer your question: Dry weight and dry mass mean the same thing in this context.

Weight = Mass x Gravity.
In typical aviation terms, in an operational sense, you don't need to worry about mass (unless you're talking about a spacecraft...)

If I need to accelerate with marginal thrust and a limited runway, I very well worry about mass.

@PeterKämpf well you may worry about mass, I will worry about weight ;)

OK, Ben, I will type more slowly. Maybe that helps. When you want to take off, you need to gain speed. Now the acceleration that acts on the aircraft's mass is not only gravity, but the acceleration gained by thrust. If thrust is low, so is acceleration. If the available runway is limited, this might very well make a takeoff impossible. Sure, a lower weight will also require less speed to fly, but during a takeoff run mass is doubly harmful.

It looks like I've offended you - sorry, that wasn't my intention. The point I was trying to make, which perhaps sp1 has better articulated, is that in typical operational circumstances, 'weight' and 'mass' are used interchangeably (even though they do have distinct meanings) when they are just trying to say weight. I'm not saying the mass of the aircraft doesn't affect anything.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Where is the spanwise flow? How does the span wise flow point the air towards the wingtip?
If you search Google for span wise flow, the first result under the pictures tab is a image of a wing with the span wise flow at the trailing edge:

(Source)
If you look at some other photos you notice others pictures that show to span wise flow at the leading edge:

(Source)
Is the flow at the leading or trailing edge?
At first I though the flow must be at the trailing edge because this would cause the airflow to point more and more towards the tip of the wing. After a little thought I realized that the air(span wise flow) should be flowing along the leading edge because that is where the air splits up in the first place.
So basically I am asking if the span wise flow is flowing along the trailing or leading edge of the wing.
How would this air cause the airflow to gradually point towards the wingtip?

@mins is right, the decomposition is a theoretical tool that we use to understand the flow. The air does not 'splits up' in to air that goes chordwise and air that goes spanwise. It's the same as when you have a diagonal force, and you split it up in horizontal and vertical forces. The force is still diagonal, but we do the split up to make the evaluation easier.

The two pictures in the question only show how to decompose the speed vector into the normal and parallel components. They are nowhere near "real" flow vectors.
First I recommend to read this answer so you get an idea how air is accelerated and decelerated when flowing around an airfoil. Note that the low pressure area over the wing will suck in more air. This answer might also be helpful because it looks closer at the boundary layer.
With a swept wing the same happens, but now the flow will not stay at one wing section but show an added sideways component. Again, air is sucked into the low pressure area, but because that area is a little to the side of the direction of motion of the wing, the flow lines will initially move towards the center of the wing on the upper side in case of sweepback.
But that cannot last: The pressure recovery over the rear part of the airfoil reduces this bending, so the flow lines become almost parallel to the direction of motion of the wing. The lower the local pressure, the more the flow lines bend inward. Higher than ambient pressure at the trailing edge or on the lower side of the wing will bend the flow lines outward accordingly.

Let the red color denote suction: It bends the flow lines (black) towards it, creating a spanwise flow component. The higher the suction (the redder the color), the bigger the bending. The cyan arrows show the flow lines inside the boundary layer. By losing some of the early acceleration to friction but being subject to the same deceleration in the region of the pressure rise, the boundary layer flow shows a stronger tendency towards the tip. To be precise: The acceleration and deceleration acts in a direction perpendicular to lines of the local relative chord. The sketch above exaggerates this bending for the sake of clarity.
Now we must look at the boundary layer to get the full picture: Here the flow speed is reduced by friction, so the result of that initial inbound acceleration is worn away gradually. The result is a twist of the local speed vectors over the thickness of the boundary layer such that the inbound component is gradually reduced the closer you get to the wing's skin. Now the deceleration in the area of pressure recovery will slow down the already decelerated air close to the wing, and again that slowing will happen in a direction perpendicular to the lines of equal chord. If this deceleration is sufficient, it will leave the boundary layer flow with an outward speed component. Such a deceleration brings the flow close to separation so it occurs only at higher angle of attack when the wing approaches stall. Note that only the boundary layer will exhibit a tipwise speed component.
Only when the flow separates, the boundary layer thickens past the separation line (blue line in the sketch above) and the spanwise flow will become significant. Separated flow is characterized by a local forward speed component, and now, past the separation line, you get a layer of air which is left with mostly a spanwise component; outboard in case of a sweptback wing and inboard in case of a forward swept wing.

EDIT
This paper presents a CFD study on the PRANDTL-D P-3C model aircraft using the OVERFLOW software. Figures 10 and 11 show the boundary layer flow on the lower and upper surface and illustrate nicely what happens. Note that this particular wing has a washout of -10° towards the tips which is essential to keep tip flow attached.

Lower surface streamlines. Only the 12° AoA case is shown because all others look quite the same.

Upper surface streamlines for angles of attack from 0° to 17.5°. The development of spanwise flow with the ever steeper pressure recovery at higher angles of attack should be evident. Note that flow direction past the separation line (the area with yellow-greenish tint) is from back to front.

I've read it four times and am still blinking. I wish I still had the books of the professor who could successfully explain concepts to freshly hungover students who had been sampling beers 1-10 in the Belgian Beer Cafe the evening before.

@Koyovis: If you could be a little more specific, I could try and improve the answer.

Where is the resulting total spanwise flow towards the tip introduced at higher angle of attack?

@Koyovis: Only the boundary layer will flow outward. In case of separation, this can be massive, but without separation this is limited to a thickening of the boundary layer on the outer wing. The outer flow will only show a tipwise speed component if local pressure is above ambient pressure.

@Koyovis: Is it any better now?

Yes thanks Peter, the arrows help very much and have stopped the blinking.

Why is the pressure lower near the wing root? Am I not understanding something?

@Crafterguy: The suction is reduced near the tips; that's all. The closer you move towards the wingtips, the more the pressure difference between upper and lower side is reduced.

The boundary layer wants to move to the tip of the wing because the trailing edge has a high pressure. Is that correct?

@Crafterguy: No, inertia makes it initially move to the tip. When separation starts there, suction is added. With separation, you are partially right: In the separated area pressure is higher than at the trailing edge. But not much. If the separation extends to the trailing edge (not uncommon), the pressure gradient is very small.

But, what force is pushing the flow to the tip before any separation or is the flow just appear to move towards the tip because the wing is at an angle. I am talking about when the wing has not stalled anywhere.

As the boundary layer slows down I would think that would bend towards the wing root because slower air has more pressure so, why does it not

@Crafterguy: No, in the boundary layer the Bernoulli law is violated by friction. Speed there is not traded for pressure, but heat. Only outside the boundary layer does Bernoulli apply. Remember, pressure-related slowing works perpendicular to the sweep lines, so a net spanwise component remains.

A reversed, outward flow in the boundary layer will cause friction in just the same way as the inward flow and still opposing the sideways flow. By that argument, the boundary layer ought to stabilise in a directly rearward direction, minimising net speed and hence net friction.

@GuyInchbald: The flow speed near the leading edge is higher and the boundary layer thickness is smaller, so here the speed gradient over the BL thickness is much steeper, causing much higher friction loss than over the rear wing.

@PeterKampf OK I have just revisited your answer and what I thought it said was not what it does say, my mistake there. My apologies. If it's OK by you I'll delete my comment and this one as a distraction.

Does this boundary layer flow (for swept wings) affects the local maximum coefficient of lift produced? If yes, why?

@Darjan: Yes. Please read this answer.

Late comment, but something I see often is that the spanwise flow will ‘build up’, meaning it will be more at the tip. Is this true? The very bottom part of your answer shows the spanwise flow on a wing and it seems to be actually less at the tip. Just wanted to see if this was a misconception.

@Wyatt Depends on what you mean with "flow". What accumulates at the tips is slow boundary layer air - the boundary layer gets massively thicker towards the tips of a swept wing at high angle of attack. Maybe you read this answer - it describes what happens.

Those pictures are valid at the leading edge, and at the trailing edge, and at all points in between on the wing area. But that is only an explanation on why wing sweep delays supersonic shock. In the pictures in the question, the real air flow is the black arrow flow marked Total, and that does not stream toward the tip. There is more to the story.
In the diagrams the black arrow marked Total is perpendicular with the fuselage. If a swept back wing is tilted at a higher angle of attack, the black arrow marked Total will start to deflect towards the wing tips. A partly stalled wing profile looks like this:

After the separation point there is low pressure. When we look at the top side of a swept back wing, the low pressure line runs along the trailing edge. Consider the red dot at the trailing edge: it sees a region with high pressure to one side (towards the root, coming from the bottom of the wing) and low pressure towards the tip, and air starts flowing towards the tip. So the streamlines look like in this picture.

So span wise flow is:

The Spanwise Component in the pictures in the question
An extra component that bends the airstream towards the wing tip.

This causes swept back wings to stall tip first, which causes a nose up pitching moment (which increases the stall) and loss of aileron control. Not desirable effects, and to be limited as much as possible.

Not sure what you mean, but have put a smaller comprehension step in the phrasing of the answer and will add more detail to the pictures when I have a chance. I find answers with too much unnecessary detail in the phrasing distracting.

But, why does the total arrow point more and more at the wingtip before the wingtip is stalled at all.

Before flow separation that hardly happens, only in the boundary layer, as @Peter Kämpf explains. At low angles of attack, for understanding purposes, it is probably best to consider 99.9 % of the air flow streaming straight through, like the black arrows in your picture. Spanwise flow is then just a vector component which explains why sweepback works in delaying compressibility effects. Real spanwise flow as in my picture is only consequential at high angles of attack, in explaining the bad stall characteristics of highly swept wings.

But consider a red dot outboard of the streamline and level with the first red dot. This one will be attracted towards the nearest low-pressure zone, which is inboard from it. So by your argument, the flow should swing wildly as it approaches the trailing edge. Also, the high pressure air below would spill up round the trailing edge, creating a local upwash of air and accompanying downforce component on the wing. I can understand this happening after the stall, but surely not before?

As some other answers have hinted, the question suffers from the misconception that the physical location of each of the vector arrows, including the small "spanwise" arrows, is intended to show the physical location on the wing where the airflow is moving in some given direction. That is not the intent at all. Any one of the vector triangles is intended to apply to the entire general area in which it is drawn. Any one of the vector triangles could have been drawn either in the manner in the top diagram, or the bottom diagram, without changing the meaning of the illustration. The smaller vectors are simply components of the larger vectors.  When we decompose a velocity vector into two or more other velocity vectors, we aren't implying that at one point in space the total velocity vector points in one direction and at another point in space the total velocity vector points in another direction.  Just like when we decompose the wind vector into "headwind" and "crosswind" components, we aren't implying that the wind is blowing in different directions at different points in space.  So both diagrams are saying the same thing-- or at least they would be if the top diagram omitted all the triangles except for one near the middle of the wing.  Similarly, it would have been equally valid to draw each set of arrows so that all three "tails" originated from a single point, or so that all three "heads" were co-located at a single point.  Arranging the arrows in a closed triangle simply makes it clear that the smaller arrows are components of the larger arrow, and the larger arrow is equal to the vector sum of the smaller arrows.1

When we speak of the "relative wind", we generally mean the free-stream relative wind created by the aircraft's motion through the airmass.  The free-stream relative wind is the direction that the airflow would take if the aircraft were not physically present to "bend", slow down, speed up, or otherwise disturb the airflow.  This is not exactly the same as the actual airflow.

So why does the upper diagram depict the "relative wind" arrows as being longer near the wing root than at the wing tip?  This makes no sense (unless the aircraft is in a very tight circle to the left!).  The length of a vector represents its magnitude, and the logical interpretation of the diagram is that the speed of the "relative wind" is greater at the wing root than at the wing tip, which is false.  Unless the arrows are only intended to represent direction and not magnitude-- which would be a poor choice of convention in an illustration intended for pilots, who are (or should be) familiar with the concept of "vectors" in the classical sense, representing a magnitude as well as a direction.

Both diagrams suffer from the error of incorrectly defining the meaning of "spanwise" and "chordwise".  "Spanwise" means perpendicular to the longitudinal axis of the aircraft, not parallel to the leading edge, trailing edge, quarter-chord line, etc.  "Chordwise" means parallel to the longitudinal axis of the aircraft, not the shortest distance from leading edge to trailing edge.  Drawn correctly, for the "relative wind" direction illustrated here, the "spanwise" arrow would have zero length, and the "chordwise" arrow would be identical to the "relative wind" arrow.

If we are interested in describing the actual airflow rather than the free-stream relative wind, the same convention applies as to the direction of "spanwise" and "chordwise".  If we re-labeled the black arrows as "airflow" rather than "relative wind", and we re-drew the red and blue component arrows correctly as described in point number 4, we would have an illustration of a swept wing that was experiencing zero spanwise airflow.

The real direction of the actual airflow over a swept wing is a complex subject that is addressed in at least one other answer to this question.

The point the diagrams seem to be trying to make is that even in the (theoretical) absence of any actual disturbance of the airflow direction by the wing-- i.e. even considering only the free-stream "relative wind"-- a swept wing would still inherently experience some airflow component that is parallel to a reference line such as the leading edge, trailing edge, or quarter-chord line.  But they don't make that point very well.  It wouldn't be correct to call that airflow component the "spanwise" airflow component, and it's confusing to make any reference to "airflow" at all when we are only considering the free-stream relative wind.

The long and short of it is that both diagrams represent typical ground-school garbage, produced by someone who likes to draw nice, bold, meaningless illustrations.

Footnotes:

Compare for example the two force vector diagrams in this related ASE answer "What does the pressure distribution over a glider's wing look like?".  The diagrams have the same meaning regardless of where the arrows are drawn, but by arranging them in a closed triangle with the head of each arrow connected to the tail of another, we make it clear that their net sum is zero.  It wouldn't change the meaning of the right-hand diagram if the red and blue arrows were arranged in a triangle on the left-hand side of the black arrow rather than on the right-hand side.


text:
What are the effects of shockwaves on normal and supersonic aircraft?
What are the effects on the speed of the plane, does it effects the body of the plane, and do shockwaves cause major damage to the plane.
I know that flying at Mach 1 is really risky, but what's the actual effect of the shockwave?

Related

Not much, really, the shock wave just streams off of the nose of the plane, and an anti-shock from the tail. They stream away from the plane and don't hit it.
Image source
The shock itself is the compression that the flying aeroplane imparts upon the unsuspecting air in front of it: disturbances travel with the speed of sound, if the aeroplane travels faster there is no warning and air gets compressed suddenly when the aeroplane arrives. This is the nature of the shock.
The amount of compression is spread out It's not a shock like a bomb explodes, only a loud noise. Flying over Mach 1 is not inherently more risky than flying subsonic. When first attempted 70 years ago there was some unexpected behaviour to iron out, but that's aviation in a nutshell for you. Once understood, a problem is only an invitation to find a solution, and that is what the aero guys did.

Wave drag and Maximum Velocity Limitations

A shockwave can limit the maximum speed of an aircraft as for each airfoil there is a critical Mach value $M_{crit}$ which is associated with the onset of a sharp rise of the drag coefficient $c_d$. This is due to the wave drag which increases as the shock grows stronger.
For this reason, in order to obtain a higher $M_{crit}$ value (i.e. increase maximum speed for same power or have the same speed for less fuel consumption) supercritical airfoils are used for transport aircrafts which employ a rather flat upper surface to delay and attenuate shock.
An other indirect effect on speed for supersonic transport aircrafts is noise regulations. For example, in the years of Concord there were incidents of broken windows near the airport, leave aside the sonic booms (have you ever heard a combat aircraft during a low-pass?). For this reason, much research is being conducted for "Low Boom" optimized supersonic aircrafts.

Transonic Buffet

When shockwave-induced flow separation happens, a flow instability may appear which is called shock buffet. This involves an oscillatory move of the shock (back and forward) accompanied with an oscillation of pressure forces on the wing which will cause vibrations to the whole body of the aircraft. This is common in fighter aircrafts in maneuvering and in early stages it can warn the pilot that he/she is approaching stall. Usually pilots are aware of the flight envelope and within it, these vibrations are not likely to cause any harm to the structure (outside of the envelope many things can happen but loss of stability and control is more likely than a major structural failure). In either case, oscillatory loads are linked to structural fatigue, and this is another issue.

See publication

“in the years of Concord there were incidents of broken windows near the airport”—there are broken windows all the time and I am sure there were some near airports where Concorde was operated, but none of them were related to the Concorde operation. While it is possible for sonic boom to break windows, it takes the jet flying 200 ft away while Concorde only went supersonic around 25,000 ft and already over the ocean.

@Jan Hudec Let the broken windows aside, the point is not the broken windows, the point is that supersonic transport aircrafts produce a sonic boom that according to noise regulations must be attenuated. Even at 25.000 ft this sonic boom produces a loud noise on the land.

Yes, it is definitely audible, and pretty annoying, from that altitude. And not only from that altitude. It is actually audible on the ground from any altitude. That is why supersonic flight is not permitted over settled land at all. Still, the talk about windows is (without further evidence) likely false and should not be mentioned.


text:
What are the differences between type B message format and EDIFACT message format?
IATA say's the standard format for transmitting passenger messages is UN/Edifact. Somewhere else it says the standard is type B format.
what are the differences between these two? Or are they a same format?

"IATA says..." can you link the document and cite the page? Also "Somewhere else it says..." Again, link the document and cite the page so we can read this in context.

There are many types of messages which are used very specifically within the aviation Industry and TypeA and Type B are the most notorious ones.
Its not the message format that differentiates them , its the characteristic message delivery and response that does.
Type A - Expect immediate application response, have timeout handling in place at the application level, delivery is not guaranteed and could expect loss of messages. EDIFACT is one of the well known industry accepted and following TYPE A message formats..
Type B - Guaranteed delivery and will make anything and everything possible to deliver the message . Does not expect an immediate response back.
Regarding the format of the messages, the type A message formats are governed by PADIS ( PASSENGER AND AIRPORT DATA INTERCHANGE STANDARDS MESSAGE STANDARDS DOCUMENT). It does differ for each applicaiton( reservation, ticketing, DCS etc). Here is a sample one for PNRGOV standards.
https://www.iata.org/iata/passenger-data-toolkit/assets/doc_library/04-pnr/PNRGOV%20EDIFACT%20Implementation%20Guide%2013_1.pdf
For TYPE B , always refer to the AIRIMP documents which governs the TYPE B mesasge formatting..
Here is a sample of the same..
ftp://ns.tais.ru/pub/doc/AIRIMP34.pdf
Thanks,
Jothi

Unfortunately, the first link is now gone, while the second is there, but contains nothing

I have searching the matter on basic definitional difference between the two type of messages. After a lot of research, my understanding of the two is given below:
1.) Type A: Transactional interchange of data between airline and the travel agent for seat reservation and ticketing. The data used is stored on single central system through data network. As you might have observed, it requires real-time query/response due to nature of job at hand i.e. reservation/ticketing, and hence, quick response. Security is not a concern though and there is high level of possibility for duplication.
2.) Type B: It is messaging between airline and service providers i.e. airports, baggage handlers, cargo handlers, immigration; and follows standardized formats as well as standardized/secure networks (SITA). Query/response is not required, however, due to nature of job, delivery of message is ensured. There are 4 levels of priorities assigned in order to ensure timely delivery.  One other matter which differentiates it from the Type-A is the cost. Due to specialized networks and organization involved during type-b messaging, costs of type-b messages is high.


text:
What is the relation of stall speed to the position of the center of pressure with respect to the centre of gravity?
Would the centre of pressure being aft or forward of the centre of gravity (CG), reduce or increase the stall speed in straight and level flight?  I know it is a weird question, I just cannot figure out if a centre of pressure being ahead of the CG (rare) would increase or decrease the stall speed, or even change it at all.

The comment written by wbeard52 is correct however this does still not explain exactly why the stall speed is increased. As an airframe stalls at an angle of attack, which will result in CLmax.

If the CG is moved to a more forward position, more downward force will be required on the horizontal stabilator to counteract the higher moment of the lift generated by the wings. This will increase the apparent weight. As (apparent) weight is increased stall speed will increase. All other parameters are fixed, in this situation.
Stall speed formula : V = √( 2 W g / ρ S CLmax )

This is a loaded question as it can possibly deal with aircraft stability, maneuverability and controllability.

Figure - 5-23 FAA Pilot Handbook of Aeronautical Knowledge
Take a look at the picture above.  For the forces to balance, there has to be one force in the opposite direction in between two other forces.  CG (center of gravity) and T (tail-down force) are used to oppose CL (center of lift).  The distance between CG and CL and CL and T are important as they provide the characteristics of stability, maneuverability and controllability.
A stall occurs when the wings reaches its critical angle of attack.  At greater weights, the wings will have to produce greater lift.  The T force is a weight that has to be countered by CL.  If CL is closer to CG, less T is required and the stall speed will decrease.  If CL is farther from CG, there is a need for more T and the stall speed will increase.  Now, as pilots, we don't talk about CL moving but changing CG.  If the airplane has a forward CG, there is a lot of length between CG and CL and a lot of T is required increasing the stall speed.  If the airplane has an aft CG, the length between CG and CL is very small and T doesn't isn't required as much.  As a result, an aft CG  reduces stall speeds.

If CL is forward of CG for an airplane like what is pictured above, the aircraft will be unstable.  There are two opposing forces on one side helping the other force cause a rotation.  Try it.
Take a pencil between your two hands.  On one hand, put your thumb and pointer finger on it.  With the other hand use your pointer finger to push on the pencil in the middle.  You should have forces just like in the picture above.  Slowly move your two pointer fingers together with the pencil in the middle.  At some point, the thumb will no longer be required to maintain stability.  Without using your thumb now, move your pointer fingers away from each other.  The pencil should rotate (very easily) out of your hand.
This is what would happen if the CL is forward of CG.

There are some airplane where the CL is forward of CG but to solve the stability issue, there is a canard which provide a nose-up force to keep the nose from rotating.  I have read, that airplanes with canards are difficult to stall due to the fact the canard will stall before the main wings.

"If CL is forward of CG […] the aircraft will be unstable." No, not necessarily. Only less stable, but still fine for regular flying.

First of all, as answered in this question, centre of pressure of the whole aircraft and centre of gravity must coincide for quasi-steady flight. Since the 1G stall speed, by definition of Part 23 and 25 regulations, is essentially quasi-steady, this holds true.
While the maximum wing lift is a fixed number for a given Mach and Reynolds number, the total aircraft CLmax can vary due to tail lift contribution. As the CG moves forward, CLmax decreases due to increased tail down force for trim. The following relationship, found in Advisory Circular 25-7C, roughly captures the influence of CG on maximum lift coefficient, which you can use to compute the stall speed:
$$C_{L_{MAX_{CG2}}} = C_{L_{MAX_{CG1}}}[1 + \frac{\bar{c}}{l_t}(CG_2- CG_1)]$$
where $\bar{c}$ is the mean aerodynamic chord (MAC), $l_t$ is the tail arm, and CG is expressed in percent MAC/100.


text:
What security regulations apply to US private airstrips?
What kind of security regulations govern the 15,000 private airstrips in the U.S.?  Do they fall under TSA jurisdiction?
I'm a freelance writer working on a story and wondering how senior Al Qaeda operative, Adnan Shukrijumah, was able to fly unhindered from Mexico to Cielo Dorado in Anthony, New Mexico.
Is this kind of laxity uncommon or par for the course at all private airstrips?

Somewhat related, but doesn't fully address the international aspect: How does security and baggage checking work in regards to a private jet?

FWIW, in the western US it's perfectly possible to land a light plane on dry lake beds or such.  Elsewhere a straight stretch of road (without adjacent powerlines :-)), or a mowed hayfield will work just fine.  I'd think the major obstacle would be radar following over the border.

Welcome to aviation.SE! Saying that private airfield security is "lax" implies that there isn't enough security, but it isn't at all clear that's a real issue. You can't secure every airfield in America any more than you can secure every parking lot or boat jetty, for both practical and legal reasons. And you also have the point that @jamesqf made: it's convenient and safer to land on a runway, but not at all necessary. Google "backcountry flying" if you want to see pilots landing in some really remote places :-)

Smuggling an individual into the U.S. by light aircraft is no more difficult than smuggling drugs by light aircraft, and a great deal of information has been published on that subject.

The problem here is not security at the field, but rather security of the airspace over the border. The flights should be tracked on radar and intercepted if they don't follow proper immigration procedures.

Respectfully, anyone who asks such fundamental questions has no business writing about the state of security at small airports. I suggest spending a few bucks on some flight lessons so you can see what goes into airport security, even at tiny remote fields. I've seen this before. An alarmist article gets "the community" in a tizzy over nothing and needless oversight is implemented to combat an imagineered threat. Also, most of those 15,000 "private" strips are actually public fields, many of which use TSA funds to implement security features like coded gates and CCTV.

Also, from a quick search, that Cielo Dorado seems to be a fly-in community, so the airport there is really no different that your driveway, or the private roads in your gated community.  If it's typical, there are probably taxiways to the houses.  What you may not understand is that a PRIVATE airport (unlike a public general aviation airport) actually is private property.  They can be as simple as a dirt strip cleared out on someone's back 40.

@acpilot all we can discern from the question is that security is relevant in some way, it's not implied that it's the topic of the story.

It's clearly implied: "Is this kind of laxity uncommon or par for the course at all private airstrips?"

Private airports have very little security and light aircraft flying within a country typically aren't required to file a flightplan and aren't required to have transponders for secondary radar. You can get in a private aircraft and fly off pretty much as easily as you can get in a car and drive off.
Private aircraft crossing international borders are subject to much stricter requirements. In the case of flights from mexico to the USA they will be required to have a transponder, to file details of the flight with the authorities in advance, to land at the first available airport of entry and to maintain a certain altitude so they remain visible on radar. At the airport of entry CBP (customs and border protection) will gather will check passports/visas, gather customs declarations and may inspect the plane. If you want to know the gory details take a look at http://www.cadds.com/bcpilots/docs/CrossBorderManual.pdf .
There is a section of airspace surrounding the USA and Canada called the North American ADIZ. All aircraft in that zone are required to identify themselves. If an aircraft is spotted in the ADIZ cannot be identified then it may be intercepted by fighter jets.
But a small plane flying low with the transponder turned off is pretty hard to keep track of on radar and if the criminal can slip undetected through the ADIZ he can easily find a quiet private airstrip to land at.

In the case of the Mexico border, the ADIZ starts at the border itself. The U.S. can't require any aircraft flying in Northern Mexico to identify itself and certainly can't intercept them there (well, not without permission from Mexico or causing an international incident, anyway.) Once they cross the border, it's fair game, though.

Even small aircraft flying low with their transponder turned off can be seen by look-down radar from the aerostats and UAVs patrolling above the U.S.-Mexico border.

"to land at the first available airport of entry" -- so if you're legitimately flying a light aircraft from Mexico home to your house in the US, basically you'd be required to land near the border, subject yourself to inspection much like anyone else crossing the border, and then resume your journey to your home airstrip?

@SteveJessop -- exactly.  You'd land at the first international airport in the US along your route, have CBP give you the all-clear, (maybe get some fuel/etal while you're at it), and then fly to your home airstrip.

@Steve Jessop: Exactly, just as if you were driving, you'd have to cross the border at one of the defined crossings, and stop for customs &c.

I don't think the "first available airport of entry" means it has to be the first beyond the border—airline, business or VIP flights don't do that. It means that your first landing must be at airport of entry (and you must have flight plan to that airport and follow it).

I can't remember where I read it now, but IIRC the rules for private flights coming from the south are much stricter than the rules for airline flights or the rules for private flights coming from other directions.

The US government would require international GA flights to land at ports of entry into the United States, designated airports which have customs personnel to meet arrivals, check passports, screen for contraband, etc. This procedure is common and said destination airports would have to be designated on the flight plan and the US dept of state notified of your intentions prior to departure.  Violation of this would alert law enforcement and, most likely, cause the plane to be intercepted by military fighters.

Some mention of the ADIZ and how it works would make this answer a lot better. This page from the FAA is also good reference.


text:
On a small/micro jet engine, can the turbine blades and vanes be made out of ceramic?
I am researching on how to make a small jet engine, and trying to figure out the materials for rotor and stator blades just after the combustion chamber so that they can hold up to the temperatures they will encounter.
I'm looking for cast repeatable shaped blades and trying to figure out what kind of forge or kiln to use.
My questions:

Should the blades vanes be nickel alloys or can they be ceramic so that dilatation won't make them interfere with adjacent parts?
Will the ceramic be too brittle?

It depends a lot on the size of the engine and what ceramic you are planning to use. Are we talking about model plane jet engine size?
There are a lot of ceramics that can easily endure the heat in a jet engine, however their tensile strength could be an issue. It could work, as miniturizing tends to put less proportional tensile stress on the components. Then again you might have a problem with foreign objects entering the engine... BUT without knowing the size of the planned engine and the planned components it's impossible to do the math if it will work or not.

It also depends on the temperatures you plan to go to.  Large commerical and military engines push the limit on temperature to get every last pound of thrust and/or every last point of fuel efficiency.  As such, they are always right up against the limit of the material.  If you are willing to sacrifice some thrust or fuel efficiency, your job will be a lot easier.

Yes, according to the journal article "Manufacturing a Ceramic Turbine Rotor for a Compact Jet Engine," Leicht et al, J. Turbomach. Aug 2023, 145(8): 081009.

As a proof of concept, an additively manufactured drop-in replacement alumina turbine rotor was produced for the JetCat P400 small-scale engine in a manner that was cost-effective, timely, and potentially scalable for production. This compact engine was used to demonstrate that a cost-effective ceramic turbine could be manufactured. At the time of publication, the desired ceramic material, silicon nitride, was not available for additive manufacturing.

The JetCat P400 weighs about 8 lb  and has a max thrust of about 90 lbf.
Similarly, Fraunhofer is making silicon nitride rotors rated for 1400ºC and 90,000 RPM for small turbines, albeit for power generation not airplane thrust.  Fraunhofer agrees that tensile strength and bending strength is a challenge for brittle ceramics.

I can't speak for all engines, but GE has been testing ceramic blades in their LEAP engines.
Here is their report on it.